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I was working on this problem:

Find the maximum likelihood estimator for the parameter $a$, in the distribution

$$f(x) =\begin{cases} 3a^3\cdot x^{−4} & \text{if } x ≥ a \\ 0 & \text{otherwise} \end{cases}$$

The likelihood function came out to be $3a^{3n}\cdot X_1\cdot X_2\cdot X_3 \cdots X_n$, which, when differentiated wrt a and equating it to $0$, gives $3n/a = 0$, which is only satisfied if $a = +\infty$ or $n = 0$, both of which don't seem right to me.

Daniele Tampieri
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Aryvd
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  • Finding the maximum likelihood estimate is not always about differentiating the likelihood and setting it equal to zero. – StubbornAtom Jun 30 '23 at 17:49
  • yeah but even when i think about it normally I think it'll still be max at infinity since there's a product x1.x2.x3..xn which is constant wrt a and a has a positive power. – Aryvd Jun 30 '23 at 17:57
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    Always consider the support of the distribution. The support is part of the density/likelihood. Your problem arises because of this oversight. – StubbornAtom Jun 30 '23 at 18:02
  • could you elaborate? i have a lot of difficulty understanding this topic in general – Aryvd Jun 30 '23 at 18:10
  • Check out https://math.stackexchange.com/q/2949033/321264 and its linked posts. – StubbornAtom Jun 30 '23 at 18:18
  • thanks i got it it now! – Aryvd Jun 30 '23 at 19:21

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